The Limit of a Sequence (Brief Summary) 1

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1 The Limit of a Sequece (Brief Summary). Defiitio. A real umber L is a it of a sequece of real umbers if every ope iterval cotaiig L cotais all but a fiite umber of terms of the sequece. 2. Claim. A sequece ca have at most oe it. Proof: Suppose L ad L be its of a sequece. The there exists a ope iterval I cotaiig L ad a ope iterval I cotaiig L ad both of these itervals cotai all but a fiite umber of terms of the sequece. If L L, I ad I ca be chose to be disjoit which (exercise) leads to a cotradictio. Thus, it must be that L = L.. Notatio. If L is the it of a sequece (a ) we write 6. Theorem. Let (a ) be a sequece. The followig are equivalet, (a) a = L (b) For every ɛ > 0, all but a fiite umber of terms of the sequece are cotaied i the iterval with ceter L ad radius ɛ. (c) ɛ > 0, N such that > N = a L < ɛ 7. Example: Let (a ) be the sequece give by a = +. Show that a =. Solutio: Let ɛ > 0 be arbitrary. We must fid a atural umber N such that 2 d(a, ) < ɛ () a = L ad we say that the sequece coverges to L or that the sequece is coverget. A sequece with o it is called diverget. 4. Remark. The defiitio of it gives precise meaig to the rather vague phrase a approaches L as approaches. This statemet does ot serve as the defiitio of it because either the ature of the approach or the cocept of are explaied. 5. Lemma. Let I is ay ope iterval cotaiig L, the withi I ca be foud a ope iterval of the form x L < ɛ wheever > N. Now d(a, ) = + = +. Moreover, sice IN is ubouded above there exists a atural umber N such that Nɛ >. It follows that: > N > ɛ ɛ > < ɛ + < ɛ d(a, ) < ɛ Philip Peace - Versio: April 0, Notice that the value of N will deped o the value of ɛ. The smaller the value of ɛ the larger N must be i order that the ope iterval determied by equatio () cotais all terms after the N th. Figures ad 4 suggest that that if ɛ = 0.2 the N = 5 works ad if ɛ = 0. the N = 9 suffices.

2 Sice N has the required property, we coclude that + = 8. Example. Let (a ) be the costat sequece give by a = 5, IN. Show that a = 5. Solutio. Let ɛ > 0. It is easy to fid N such that a 5 < ɛ wheever > N. Let N = 24 (ay other atural umber will do). Sice a 5 = 0, for all IN it is clearly true that a 5 < ɛ wheever > 24. It follows that a = More geerally, if a = c for all IN the a = c. a a 2 a a 4 a Figure. Iitial Terms of the Sequece a = Figure 2. Graph of the sequece a = + L =.2 L =.0 L 0.2 = Figure. > 5 a <

3 L = L + 0. =. L 0. = Figure 4. > 9 a < Example: Let (a ) be the sequece give by a =, >. Show usig the defiitio of the it of a se- quece that a = 0. Solutio. Let ɛ > 0 Notice (exercise) that < 2 for all > 2. Sice IN is ubouded, there exists a atural umber N > such that Nɛ > 2. It follows that: > N ɛ > 2 2 < ɛ Sice < 2, it follows that > N < ɛ 0 < ɛ which proves that d(a, 0) < ɛ =. The ext theorem reduces the calculatio of complex its to simpler oes. 2. Theorem. Let (a ) ad (b ) be sequeces. a = L ad b = M the: (a) (a ± b ) = L ± M (b) (a b ) = L M (c) If M 0 ad b 0 for all the a b = L M.. Example. Fid the it of the sequece (a ) give by a = / Solutio. Notice a = x y where x = / ad y =, IN. Now, it is easy to show that x = 0 ad y =. Usig part (b) of the above theorem yields a = 0 = 0 4. Example. Fid the it of the sequece (a ) give by a = 2. Solutio. Let x = /. The a = x x If

4 Sice x = 0. a = x x = 0 0 = Usig mathematical iductio (exercise) it is ow easy to show that p = 0 for ay atural umber p > Example. Fid the it of the sequece x = Solutio. But ad Hece 2 x = = ( + ) = + = + 0 = ( ) + = x = =. = 0 + = ( + ) ( ) Defiitio. A sequece (a ) is bouded if there exists a umber M such that a M for all IN. A sequece which is ot bouded is called ubouded. 8. Example. Cosider the sequece a = +. Sice a for all IN, this sequece is bouded. 9. Claim. A coverget sequece is bouded. Proof. Let (a ) be a coverget sequece with it L. The there exists a umber N such the ope iterval (L, L+) cotais all terms a with > N. It follows that all terms of a are less tha or equal i absolute value to the maximum M of the set { a, a 2,..., a N, L +, L } (Exercise: Fid a couterexample to show that the coverse of this result is false.) 20. Corollary. The sequece a = has o it. Proof. Notice that the sequece (a ) is ubouded. If it had a it it would be bouded. 2. Sadwich Theorem Let (a ), (b ) ad (c ) be sequeces. If a b c for all IN ad the seqeces (a ) ad (c ) have the same it L the b = L cos 22. Example. Show = 0 Solutio. Notice that for all 0 cos The result follows by applyig the sadwich theorem. 4

5 2. Claim. The followig are equivalet: (a) a = 0. (b) a = 0. Proof. Exercise Hit: (Notice that d( a, 0) = d(a, 0) 24. Corollary. cos = Lemma. (Beroull s Iequality) If x > the ( + x) ( + x) (b) o-decreasig if for all, m IN m x x m (c) decreasig if for all, m IN < m x > x m (d) o-icreasig if for all, m IN m x x m Warig: As show i the Ve diagram, the terms o-icreasig ad icreasig are ot mutually exclusive. Proof. By iductio (Exercise). 26. Claim. If a > the a is ubouded above. Proof. If a > the a = + x where x = a. Notice x > 0. By Beroulli s iequality a = ( + x) > + x Sice IN is ubouded above, give M arbitrary, there exists N IN such that for all > N, x > M. Hece, for all > N a > M provig that the sequece (a ) is ubouded above. 27. Claim. If 0 < a < the a = 0 Proof. Sice >, it follows by the a previous claim that the sequece ( ) a is ubouded. It follows that if ɛ > 0 there ( exists N such that for all > N, ) a >. This implies ɛ that a 0 < ɛ for all > N. Hece a = Defiitio. A sequece (x ) i IR is: (a) icreasig if for all, m IN < m x < x m Icreasig ic A No-decreasig Decreasig dec C No-icreasig 29. Dedekid (Completeess) Axiom. Every o-empty subset of IR which is bouded above has a supremum (least upper boud). 0. Claim. A sequece α = (a ) of real umbers which is both o-decreasig bouded coverges to the least upper boud of its set of terms. Proof. Let L = sup{a : IN} ad ɛ > 0. The there exists N such that L ɛ < a N L (otherwise L ɛ would be a smaller boud). Sice α is o-decreasig N L ɛ < a L Hece α coverges to L. 5

6 . Corollary. A o-decreasig sequece α = (a ) of real umbers either coverges or is ubouded above. 2. Defiitio. Let (x ) be a real sequece. A atural umber m is called: (a) A peak poit of (x ) if x m x for all m. (b) A p-blocker if m > p ad x m x p. Bolzao Weierstrass Theorem. A bouded sequece has a coverget subsequece. Proof. (See Spivak). Let m < m 2, <..., be the set of peak poits. If this set is ifiite the the bouded o icreasig subsequece x m x m2 x mk... must coverge (to its ifimum). If, o the other had, the set of peak poits is fiite, there exists a last peak poit m i which case there exists a coverget subsequece of m-blockers defied recursively as follows: let q = m + ad, for each k, let q k+ be the miimum blocker of q k. The the subsequece x q x q2 x qk... is o decreasig ad bouded below so coverges. 4. Exercise. I the proof of BW above, explai why the miimum at each stage exists. 5. Defiitio. A real sequece (a ) is Cauchy if for every ɛ > 0 there exists N such that, m > N a a m < ɛ. 6. Claim. Cauchy sequeces are bouded. Proof. Let ɛ > 0. There exists N such that for m, N, a m a < ɛ. The, by the triagle iequality a m a a m a < ɛ for all m, N. Takig = N. The a m a N < ɛ for all m N. It follows easly that (a m ) is bouded by ± max { a 0, a,..., a N, a N, ɛ + a N } 7. Claim [Uses BW]. Every Cauchy sequece i IR coverges. Proof. Let (a ) be a Cauchy sequece of real umbers. The (a ) is bouded so by BW has a coverget subsequece (a k ). Let L be the it of this subsequece. The there exist N such that k > N a k L < ɛ/2 ad there exists M such that, k > M a a k < ɛ/2 The, for all > max{n, M} a L a a k + a k L < ɛ 8. Corollary. Dedekid completeess implies Cauchy completeess 6

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